Geometric inequality: $2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$ Suppose $a$, $b$, and $c$ are the lengths of the sides of a triangle, and $R$ and $r$ are its circumradius and inradius respectively. How can one prove the following inequality? $$2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$$
 A: We know that 
$$\begin{align*}
a+b+c&=2s,\\
ab+bc+ca&=s^2+r^2+4rR,
\end{align*}$$
where $s$ is semiperimeter, $r$ is inradius and $R$ is circumradius; 
see e.g., Manfrino, Ortega, Delgado - Inequalities: A Mathematical Olympiad Approach, Section 2.5.
Using these inequalities we can rewrite the LHS as
$$
\begin{align*}
2(r^2+4rR) &= 2(ab+bc+ca-s^2) \\
 &= 2(ab+bc+ca)-\frac{(a+b+c)^2}2 \\
 &= ab+bc+ca-\frac{a^2+b^2+c^2}2.
\end{align*}$$
After this the inequality we want to prove becomes
$$ab+bc+ca\le a^2+b^2+c^2,$$
which follows easily from AM-GM inequality. (Simply add the inequalities $ab\le \frac{a^2+b^2}2$, $ac\le \frac{a^2+c^2}2$ and $bc\le \frac{b^2+c^2}2$. )  See also this question for more proofs of the last inequality.

For the sake of completeness, let me copy here the proofs of these equalities as they are given in the book I mentioned above. (I hope that such a short excerpt still qualifies as a fair use. I tried to find a proof of them online, but I did not succeed.)

$$
\begin{align*}
a + b + c &= 2s, \tag{2.5}\\
ab + bc + ca &= s^2 + r^2 + 4rR, \tag{2.6}\\
abc &= 4Rrs. \tag{2.7}
\end{align*}
$$
  The first is the definition of $s$ and the third follows from the fact that the area of
  the triangle is $\frac{abc}{4R}=rs$. Using Heron’s formula for the area of a triangle, we have
  the relationship $s(s - a)(s - b)(s - c) = r^2s^2$, hence
  $$s^3 - (a + b + c)s^2 + (ab + bc + ca)s - abc = r^2s.$$
  If we substitute $(2.5)$ and $(2.7)$ in this equality, after simplifying we get that
  $$ ab + bc + ca = s^2 + r^2 + 4Rr.$$ 

A: Consider the triangle formed by the circumcenter $O$,$A$, and $B$. Using the law of cosines on the angle $\alpha$ formed by $OA$ and $OB$ we obtain, where $R$ is the circumradius:
$$a^2=2R^2-2R^2\cos \alpha\Rightarrow \frac{a^2}{2}=R^2-R^2\cos \alpha$$
Clearly, analogous identities hold for the other two sides, so substituting these into the right side of the inequality, we now wish to prove 
$$R^2(3-\cos\alpha-\cos\beta-\cos\gamma)\ge 2r^2+8Rr$$
It is not hard to see (exercise left to reader) that $-\cos\alpha-\cos\beta-\cos\gamma\ge 1$ for all angles in the domain of the problem, because $\alpha+\gamma+\beta=2\pi$ and they are all between $0$ and $\pi$. So we will prove the stronger inequality
$$4R^2\ge2r^2+8Rr\Rightarrow 2R^2\ge r^2+4Rr$$
This is equivalent to $$6R^2\ge r^2 + 4Rr +4R^2=(r+2R)^2$$
So it remains to prove $\sqrt 6 R\ge r+ 2R$. Unfortunately, this is not true in general. In general, we only have $2R\ge r$, which is not strong enough. But in the cases where the contribution to coefficient on the left hand side from the sum of the cosines is small enough that Euler's inequality doesn't apply, one of the angles is very, very small, the triangle becomes very skinny, and $r$ is clearly small enough that the inequality holds.
A: We need to prove that
$$2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$$ or
$$2\cdot\frac{4S^2}{(a+b+c)^2}+8\cdot\frac{abc}{4S}\cdot\frac{2S}{a+b+c}\leq\frac{a^2+b^2+c^2}{2}$$ or
$$\frac{\prod\limits_{cyc}(a+b-c)}{2(a+b+c)}+\frac{4abc}{a+b+c}\leq\frac{a^2+b^2+c^2}{2}$$ or
$$(a^2+b^2+c^2)(a+b+c)-8abc+\sum_{cyc}\left(a^3-a^2b-a^2c+\frac{2}{3}abc\right)\geq0$$ or
$$a^3+b^3+c^3-3abc\geq0,$$
which is AM-GM.
Done!
